Can anyone prove that .999999 repeating, is equal to 1?

Question

There is a mathmaticle proof to do it.

Answer 1

rgetso is correct... there are tons of proofs in Y! Answers to this question. (I have a couple of "Best Answers," myself, for the very same question.)

Yes, 0.999999... recurring, repeating forever, DOES equal 1. It's merely another form of writing 1, such as 3/3.

What all do you need for proof? Here's an easy one:

1/3 = 0.333333... repeating.
2/3 = 0.666666... repeating.
Add the two lines and you get
3/3 = 0.999999... repeating. 3/3 = 1, so 0.99999... repeating = 1.

Not formal enough for a proof? Here is algebraic one:

Let x = 0.99999... repeating.
Multiply both sides by ten.
10x = 9.99999... repeating. [On the right, multiplying by 10 is a simple move of the decimal point.]
Now find the difference of these two lines by subtracting the top from the bottom.
10x - x = 9.99999... - 0.99999...
9x = 9
x = 1
If x = 0.99999... repeating (given) and x = 1 (solved), then
0.99999... repeating = 1.

How about infinite sums?

0.99999... repeating = 9(0.1)¹ + 9(0.1)² + 9(0.1)³ + ...
This is an infinite sum of ∑ 9· (0.1)^k, where k goes from 1 to ∞.
∑ = a / (1 - r), where a is the first term in the series and r is the common ratio of successive terms.
0.99999... = ∑ = 0.9 / (1 - 0.1) = 0.9 / 0.9 = 1.

People who can't believe 0.99999... repeating = 1 are the same people to whom Zeno wrote over 2400 years ago, believing Achilles will never catch the tortoise in the race. http://en.wikipedia.org/wiki/Zeno's_paradoxes#Achilles_and_the_tortoise

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Note to A Guy: I showed three proofs. You've only disputed one. The other two work very nicely, Zeno-lover! :P

2nd Note: You're trying to tell me that it's impossible to take an infinite sum?!?!? It's Algebra 2 stuff! You've also neglected to refute this very simple, elegant proof that uses only 2300 year old knowledge... equals added to equals are equal and the transitive property:

1/3 = 0.333333... repeating.
2/3 = 0.666666... repeating.
Add the two lines and you get
3/3 = 0.999999... repeating. 3/3 = 1, so 0.99999... repeating = 1.

And you can find all sorts of cool characters (like ∞ and ∑) by grabbing your Character Map. (Start, All Programs, Accessories, System Tools, Character Map.)

3rd Note: My apologies for the insults, they were unwarranted and I've edited them out [except Zeno-lover ;-) ] I do have to admit a lot of people have a difficult time philosphically with the use of infintesimals. (An infintesimal being a theoretically infinitely small number that's not equal to zero, such as the difference of 1 - 0.99999... repeating forever.)

The reason the "dropped digit" works in the algebraic proof is because the "infinite number" of 9's in x is of the same cardinality or class of infinity as the "infinite number" of 9's in 10x. (See references to Georg Cantor http://en.wikipedia.org/wiki/Georg_Cantor or aleph numbers http://en.wikipedia.org/wiki/Aleph_number .)

One more popular proof these days that 0.99999... (repeating forever) = 1 deals with the notion of betweenness. 0.999 is not equal to 0.99999... (repeating forever) because there exists a number that lies between them (for example, 0.9991). 0.99999... (repeating forever) DOES equal one, though, because there is no number that exists that's between the two.

Note to arvind_vyas:
Nobody likes to deal with infintesimals! As noted a couple paragraphs above, using them to prove or disprove anything in math is borderline unethical. That's why Weierstrass re-invented the entire calculus using limits... to get rid of the pesky infintesimals brought into play by Leibnitz and Newton.

Note to rgetso:
¥öú´®ê \/\/è£çømë.

4th note to A Guy:
There is a number between 0.99999...8 and 1, because the 8 on the end forces the number of 9's in the string to be finite. The number between them is 0.99999...9 (with the ellipsis representing the same finite number of 9's). If the number of 9's was infinite, there would be no place to put an 8 on the end.

Note to Tom:
I looked in wikipedia as you suggested. There I found seven proofs that 0.99999... = 1. http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1

The article closes, "This topic provokes interest far beyond its minor status within mathematics. For example, in the newsgroup sci.math, devoted to discussion of general mathematics, statistics show over one thousand postings related to this proof; and it is one of the questions answered in its FAQ. It is also quite common in other forums of an elementary nature. One reason might be that people encounter it at a time when they are young and curious, and the usual explanations seem unconvincing. Another is that, like many such magnets, the statement of the proposition is elementary, but the proof is not.

"Professor David Tall has gone so far as to study characteristics of teaching and cognition that might lead to some of the misunderstandings he has encountered in his college students. Interviewing his students to determine why the vast majority initially rejected the equality, he found that 'students continued to conceive of 0.999... as a sequence of numbers getting closer and closer to 1 and not a fixed value, because you haven’t specified how many places there are or it is the nearest possible decimal below 1.'

"Many Internet message boards contain frequent debates over this theorem since some participants reject it."

Note to Jay T:
What is it about an infinitely long decimal that you don't understand? Your so-called argument uses only finite numbers of 9's. If you have a string of 9's that's infinitely long, you can't add another 9 on the end of it. If you could, your original string wasn't infinitely long to begin with!

No, 0.999999999999 is not 1,
0.9999999999999999999999 is not 1,
but 0.99999... (repeating forever) is indeed another form of writing the number 1, as easily as 3/3 is in the proofs shown above your answer, and as easily as I can comfortably eat my pudding!

Answer 2

Before you are ready to accept a proof in mathematics, you need to have a clear understanding of what the terms involved mean.

What I want you to focus on is, what *is* .9999...? Or any decimal expansion, like 3.1415... or whatever.

A number doesn't change. It's not 0.9 one moment and then 0.99 the next, as if some outside person is changing it by contemplating it. It's stuck somewhere on the number line, no matter how you describe it.

And if I ask, what is the difference between 1 and 0.9999...., it's going to have to be a fixed number too, not some "infinitely small" thing. There has to be an answer; you can always subtract numbers. And it's easy to see, as many have pointed out, that the difference, if positive, would have to be smaller than any other positive number. If not positive, it must be zero.

Can it be that there is a positive number smaller than any other positive number? No, then it would be smaller than itself (or half itself). Contradiction.

Here's another perspective: do you believe that 1/2 + 1/4 + 1/8 + 1/16 + ...=1? (Classic Achilles & the Hare problem.)
That's the same as saying 0.11111.... in binary is equal to 1. Your problem is the same idea base ten.

Answer 3

rt111 has the most famous proof of this, but .999999 repeating doesn't actually EVER equal 1. It always equals .999999 repeating, and nothing else! The reason being is that the "proof" that rt offered uses an incorrect definition of infinity (IE. the repeating of the .999999 ad infinitum). Simply put, his "proof" works only for .999999 repeating because the question of the lost digit between X and 10X is "answered" because it repeats! However, when working with infinity you can't simply make such assumptions that a lost digit is insignificant because it will be replaced by the infinite other digits. There will always be a difference in the number of digits that each number consists of, so they never truly operate in the way the equation makes it seem. In other words, while superficially it seems correct that 10X-X=1, it doesn't because there is an extra digit in X as opposed 10X. Basically, 10X-X=9X is infinitesimally close to 9, but not actually 9!

Does that make sense?

And Louise, while I appreciate your annoyance with illogical attempts to dispute this question, my answer is entirely logical, and I should add, CORRECT! I did enjoy the explanation of the infinite and it proves why we say that .999999=1, but it still fails to acknowledge the problem of the dropped digit. (PS how did you do sigma and infinity notation?)

Listen, you are the math teacher so your knowledge is severely better than mine. I haven't disputed your fractional proof because it is indisputable. And, while I obviously don't disagree with the idea of infinity sums, ever since I was exposed to it in Calculus 1, I've always had a problem with the idea of the dropped digit in just this problem. In fact, as I'm sure you can imagine, my math teacher did this problem at the time (8 years ago, now), but even then it made no sense because of the dropped digit problem! He couldn't explain it to me then but to simply say "it's the definition", but my question was always WHY? So, since you are obviously better at it than he, and certainly I, why? Why do we just forget the dropped digit? Why isn't X always one digit behind 10X? And I can't accept the "we simply moved the decimal so there are still the same number of digits" explanation because it seems the decimal is always static.

EDIT

Excellent. Cantor's cardinality was the basis for my question in the first place. I thought it shifted N to a different cardinality because of the dropped digit, but I'm wrong in said supposition. Now I recognize the error in my ways. As far as the more "modern" solution that .999999 must equal one because there is no average of the two numbers and no number between them seems problematic to me, too, though. Pardon my lack math knowledge, but couldn't the same argument be said about .999999........8 equaling .999999.... since there is no average or number between the two? Thus, couldn't the statement .99999999..........8 = 1 be true?

Answer 4

Yes, lots of the same problems are asked every day. But, youngster's don't always check everything.

So, let's be patient.

This is a very nice problem. Between any two different numbers a and b, there are infinitely many numbers. Trying to find one between .99999... and 1.0000... is a problem so they must be the same: since there isn't a real number between .999... and 1 they must be equal.

Decminal representations are not unique. In the case of .999..... you have a geometric series with a=9/10 and r=1/10 so it converges to

a/(1-r)=(9/10)/(1-(1/10))=(9/10)/(9/10)=1.

The above is a bit technical. I prefer the fact that between any two distinct real numbers a and b there are infinitely many real numbers that are different from a and b. In the case of .999.... and 1.000... there aren't any, so they must be the same.

Decimal representation is not unique; reduced fraction representation is unique. It is important for secondary teachers to stress to the difference between reduced fractions and decimal representation of numbers. Unfortunately, I don't think with all the tests that are given each year, they have the time to explain that.

I apologize for adding political views!

Answer 5

one equals a limit as follows;
1 = lim of the sum as n aproaches infinity = 9/10 + 9/100+ 9/1000 +9/10000+......9/(10^n)

That means no matter how much the series is added it will never exceed(1) the number one.
If we substract 1 form the series for a value of n the difference would be 1 x10^-(n-1).
and if we add the two series together we obtain 1.

Answer 6

If one more person asks this obvious question, then the whole stinking Internet is going to self destruct! Or maybe it's just me that will self destruct.

One last time...

If we let N = 0.999999…,
then 10N = 9.999999….

10N - N = 9
9N = 9
N = 1

Therefore, 1 = .999999….

If you will do a search of Yahoo! Answers, then you will find that this question is asked about once every 4 or 5 days. Some of the solutions are quite interesting in their accuracy and creativity. Get to searching.

◄Update► I've created a firestorm of controversy. Ahhh, I feel better now. I prefer Louise's ∑ proof because it is so simple and to the point. However, most folks probably won't understand it if they don't understand that 0.99999…=1.

Now, for all those folks that don't believe 0.99999…=1. Let's suppose that 0.99999…≠1. Then 9/9≠1 (DOH! I just used what I was trying to disprove in my argument). Great! Now our whole system of arithmetic just fell apart. Thanks alot. ☺

Louise, thanks for reminding me about the character map! ♥♫ Your mention of betweeness reminded me of [Richard] Dedekind's Cut. It's all starting to come back to me now.

Answer 7

0.99999999999999999999999

is not 1.0 and any 'proofs' to the contrary are mathematically invalid.

Such a proof is impossible - unless flawed.

Adding another 9 to the sequence only makes it closer to 1, but never EQUAL to 1 exactly.

Understanding this is very, very critical to properly understanding mathematical concepts correctly, especially asymptotic concepts.

A calculator will usually round it off to 1.0, which is OK in most cases because for most practical purposes, 0.99999999999999999999999 can be replaced by 1.0 and any error is insignificant.

But when it comes to proper, formal mathematics, it should never be equated to 1.0 to be strictly correct because it simply is not true any more than 0.33333333333333333333333 is exactly 1/3

I'm a little disappointed at some of the math 'experts' here. Shame, shame, shame! No pudding for you tonight.

Answer 8

0.999... is not equal to 1.

Every one of these so-called proofs (they are all false) have been refuted in the Wikipedia arguments page. Again, you are encouraged to go and read it. Nothing that has been said here has not already been discussed in the Arguments page. I know because I followed all the arguments. The only one proof that comes close to being convincing is one by a guy called Rasmus. However, it was also shown to be flawed. It's not a wise idea asking the question here again, rather go and read up on it in the Arguments page of Wikipedia. I believe everything is still there - you can also read much of it in the Wiki archives.

Answer 9

Let x = .999999...
10 x = 9.999999...

9x = 10x -x = 9.999...-.999... = 9

9x = 9
x = 1

Note to "A Guy":
Either you accept that .9999... represents a real number or you don't.
If you accept that .9999... represents a number, then it has to follow the same rules of arithmetic as all the other real numbers.
The steps above show that .9999... = 1.

Answer 10

As .999999 repeats, it approaches the value of 1, but never reaches that limit (i.e. never actually can become 1). Its normally a rounding thing.